# If the lines $\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}$  are perpendicular, find the value of $\lambda.$ Hence find whether the lines are intersecting or not.

The equation of the given lines are:

$\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}$

If these lines are perpendicular, then

$a_1a_2+b_1b_2+c_1c_2=0$

$\Rightarrow -3 \times 3\lambda +2\lambda \times 2+2\times \left ( -5 \right )=0$

$\Rightarrow -9\lambda +4\lambda -10=0$

$\Rightarrow -5\lambda =10$

$\lambda =-2$

Now the lines are

$\frac{x-1}{-3}=\frac{y-2}{2\lambda}=\frac{z-3}{2}\: \: and\: \: \frac{x-1}{3\lambda}=\frac{y-1}{2}=\frac{z-6}{-5}$The coordinates of any point on first-line are given by:

$\frac{x-1}{-3}=\frac{y-2}{-4}=\frac{z-3}{2}=\alpha$

or $x-1=-3\alpha \Rightarrow x=-3\alpha +1$

$y-2=-4\alpha \Rightarrow y=-4\alpha +2$

$z-3=2\alpha \Rightarrow z=2\alpha +3$

So, coordinates of any point on this line are $(-3\alpha+1, 4\alpha+2, 2\alpha+3)$

The coordinates of any point on the second line are given by:

$\frac{x-1}{-6}=\frac{y-1}{2}=\frac{z-6}{-5}=\beta$

or $x-1=-6\beta \Rightarrow x=-6\beta +1$

$y-1=2\beta \Rightarrow y=2\beta+1$ $z-6=-5\beta \Rightarrow z=-5\beta+6$

So, coordinates of any point on second line are $\left ( -6\beta+1, 2\beta+1, -5\beta+6 \right )$

If lines intersect then they have a common point. So far some values of $\alpha \: \: and\: \: \beta$ we have

$-3\alpha+1= -6 \beta +1$$\Rightarrow -3\alpha= -6 \beta$

$\Rightarrow \alpha= 2 \beta$

and $\Rightarrow -4\alpha+2= 2 \beta +1$

$\Rightarrow -4\alpha+1= 2 \beta$

On solving, we have

$\alpha =\frac{1}{5}\: \: \: and\: \: \beta=\frac{1}{10 }$

The values of $\alpha \: and\: \beta$ do not satisfy the third equation.

Hence, lines do not intersect with each other.

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